Showing that $a_n$ is a contracting sequence I have a sequence defined by $a_1=\frac{1}{2}$ and $a_{n+1}=\frac{1}{2}(2+a_n-a_n^2)=\frac{1}{2}(\frac{9}{4}-(a_n-\frac{1}{2})^2)$
I have shown that $\frac{1}{2}\le a_n\le\frac{9}{8}$ for all $n$ and now need to show that $a_n$ is a contracting sequence. I know the definition of a contracting sequence to be there exists $0\le\lambda<1$ such that $|a_{n+2}-a_{n+1}|\le\lambda|a_{n+1}-a_n|$,  $\forall n$ in $N$. 
Then I considered $a_{n+2}-a_{n+1}$ and eventually rearranged to $\frac{1}{2}((a_n-1/2)^2-(a_{n+1}-1/2)^2)$ and then also $\frac{1}{2}(a_n^2-a_n-a_{n+1}^2+a_{n+1})$. 
I am unsure what to do from here, I know I somehow need to compare a form of this expression to $a_{n+1}-a_n$ but how? Am I on the right lines? 
 A: So simplify further and factor the difference: 
$a_{n+2}-a_{n+1}= \frac{1}{2}(a_n^2-a_n-a_{n+1}^2+a_{n+1}) = 
\frac{1}{2}(a_{n+1}-a_n-(a_{n+1}^2-a_n^2)) = 
\frac{1}{2}(a_{n+1}-a_n-(a_{n+1}-a_n)(a_{n+1}+a_n)) = 
\frac{1}{2}(a_{n+1}-a_n)(1-(a_{n+1}+a_n))$. 
Note $a_{n+1}-1=\frac{a_n(1-a_n)}2$. Also, if $a>1$ then $a-a^2<0$ so $\frac12(2+a-a^2)<\frac12\cdot2=1$, and similarly if $0<a<1$ then $\frac12(2+a-a^2)>1$. Since $a_1=\frac12<1$ it follows that $a_2=\frac98>1$ so $a_3<1$. 
I will use this to prove by induction that $|a_{n+1}-1|\le\frac12$, i.e. $\frac12\le a_n\le\frac32$, and that $a_n>1$ if $n$ is even, and $a_n<1$ if $n$ is odd. True for $n=1$. If $\frac12\le a_n\le1$ (when $n$ is odd) then $0\le1-a_n\le\frac12$ so $|a_{n+1}-1|=\frac{a_n(1-a_n)}2\le\frac14$. 
If $1\le a_n\le\frac32$ then $|1-a_n|\le\frac12$ so $|a_{n+1}-1|=\frac{a_n|1-a_n|}2\le\frac38$. 
The above proves $\frac12\le a_n\le\frac32$ and that the $a_n$ alternate on different sides of $1$. It follows that 
$|a_{n+1}-a_n|\le1$ and $\frac32\le a_{n+1}+a_n\le\frac52$, so $|1-(a_{n+1}+a_n)|\le\frac32$. Note $a_2-a_1=\frac58$. Prove by induction that both $|a_{n+1}-a_n|\le\frac58$ and $|a_{n+2}-a_{n+1}|\le\frac34\cdot|a_{n+1}-a_n|$ (so $\lambda=\frac34$ works). 
We have 
$|a_{n+2}-a_{n+1}|=\frac{1}{2}|(a_{n+1}-a_n)(1-(a_{n+1}+a_n))|\le$ 
$\le\frac12|(a_{n+1}-a_n)|\frac32 = \frac34|(a_{n+1}-a_n)|\le\frac34\cdot\frac58
\le \frac58$. 
This could all be written in a better way and perhaps simplified, but I think it is right. 
The following older answer applies to the original statement, 
which was that $a_1=1$ (which was later changed to $a_1=\frac12$). 
I get that $a_2=1$, so $a_n=1$, does that help? $|a_{n+1}-a_n|=0$. 
A: $$\frac12(a_n^2-a_n-a_{n+1}^2+a_{n+1})=\frac12(a_n-a_{n+1})(a_n+a_{n+1}-1)$$
Can you show that $|\frac12(a_n+a_{n+1}-1)|<1$?
